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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Level
MARK SCHEME for the May/June 2012 question paper
for the guidance of teachers
9231 FURTHER MATHEMATICS
9231/11 Paper 1, maximum raw mark 100
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.
Mark schemes must be read in conjunction with the question papers and the report on the examination.
• Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
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Page 2 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
Mark Scheme Notes
Marks are of the following three types:
M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.
A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.
Accuracy marks cannot be given unless the associated method mark is earned (or implied).
B Mark for a correct result or statement independent of method marks.
• When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given.
• The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working.
• Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2.
The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.
• Wrong or missing units in an answer should not lead to the loss of a mark unless the
scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f.,
or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
The following abbreviations may be used in a mark scheme or used on the scripts:
AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that
the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely
clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error
is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently
accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a
case where some standard marking practice is to be varied in the light of a particular circumstance)
Penalties
MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or
part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting.
PA –1 This is deducted from A or B marks in the case of premature approximation. The
PA –1 penalty is usually discussed at the meeting.
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
Qu
No
Commentary Solution Marks Part
Mark
Total
1 States ∑α and ∑αβ
Uses formula for correctly. Uses formula for ∑ 3
α
to obtain result.
∑ = 7α ∑ = 2αβ
∑ =×−= 4522722
α
∑ ∑ ∑ +−= 92723
ααα
= 315-14+9 = 310
B1
B1
M1
A1A1
2
3 [5]
2 (States proposition.) Proves base case. States inductive hypothesis. Proves inductive step.
States conclusion.
(Pn: 4n > 2n + 3n)
Let n = 2, 16 > 4 + 9⇒P2 is true. Assume Pk is true ⇒4k > 2k + 3k
kkkkkk 3.42.4)32(44.44 1+=+>=
+
11
323.32.2++
+=+>kkkk
∴Pk⇒Pk+1
Hence result true, by PMI, for all integers n ≥ 2.
B1
B1
M1
A1
A1 (CWO) 5 [5]
3 Proves initial result.
Sets up method of differences. Shows cancellation to get result. States sum to infinity.
f(r – 1) – f(r) = )2)(1(
1
)1(
1
++
−
+ rrrr
)2)(1(
2
)2)(1(
2
++
=
++
−+=
rrrrrr
rr
(AG)
×−
×=
++∑
32
1
21
1
2
1
)2)(1(
1
1
n
rrr
……
++−
++
)2)(1(
1
)1(
1
2
1
nnnn
++−=
)2)(1(
1
2
1
4
1
nn (OE)
4
1
)2)(1(
1
1
=
++
∴∑∞
rrr
M1
A1
M1A1
A1
A1√
2
3
1 [6]
‘Non hence’ method
for last two parts
i.e. penalty of 1 mark.
)2(2
1
)1(
1
2
1
)2)(1(
1
+
+
+
−=
−+ rrrrrr
⇒⋅⋅⋅⇒
)2(2
1
)1(
1
)1(2
1...
4
1
2
1
2
1
+
+
+
−
+
++−
nnn
++−=
)2)(1(
1
2
1
4
1
nn
(OE)
4
1
)2)(1(
1
1
=
++
∴∑∞
rrr
(M1)
(A1)
(A1)
(A1√)
(3)
(1)
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
4 Draws sketch of C.
Uses θβ
αd
2
1
2∫ r
Uses double angle formula.
Integrates and obtains area.
Finds areas.
Shows (4,0) and (0,π) lie on C. Correct shape. (Full cardioid is B1 unless clear evidence of plotting up to 2π or –π to π.)
∫ ++
π
θθθ
0
2 d)cos4cos84(2
1
∫ ++=
π
θθθ
0
2 d)cos2cos42(
∫ ++=
π
θθθ
0
d)2coscos43(
πθ
θθ
π
32
2sinsin43
0
=
++=
(A1 for correct integral)
712.45
cos5
sin5
sin45
3=++
ππππ
713.4712.43 =−π
B1 B1
M1
M1
A1A1 CWO
M1A1
A1
2
4
3 [9]
5 Identifies matrices P and D.
Finds inverse of P.
Uses appropriate result to obtain A. (First mark can be implied by correct working.)
P =
−
−−
−
532
311
210
D =
−
200
010
001
Det P =1
P–1 = Adj P =
121
241
5114
P–1
AP = D ⇒ A = PDP–1
A =
−
−
−−−
−
121
241
5114
200
010
001
532
311
210
=
−−−
−
121
241
5114
1032
611
410
=
−−−
265421
132711
243
B1B1
B1
M1A1
M1
M1A1√
A1
9 [9]
5 Alternative Approach:
A =
ihg
fed
cba
Use of Ae = λe
Obtains 3 sets of 3 linear equations: One set Other two sets Solves one set Solves other sets
(M1)
(M1A1) (A1A1)
(M1A1) (A1A1) (9) [9]
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
6 Obtains fifth roots of unity by de M’s Thm.
Rewrites
and factorises.
Isolates z. Obtains purely imaginary denominator
and obtains result.
1 = cos(2kπ) + i sin(2kπ) k = 0, 1, 2, 3, 4.
=
5
2 πθ
k k = 0, 1, 2, 3, 4.
11
1)1(
)1(
5
5
5
55 =
+⇒=
+⇒=+
z
z
z
z
zz
=+⇒
=
+
5
2 cis 1
5
2 cis
1 ππ kzz
k
z
z
15
2cis1 −=
−⇒
πkz
−
−
−−
=
−
−=⇒
5cis
5cis
5cis
5
2cis1
1
ππ
π
π kk
k
kz
,3 ,2 ,1 5
coti 2
1
2
1
5sin i 2
5sin i
5cos
=
π+−=
π−
π+
π−
= kk
k
kk
(Alternatively for the above three marks – rationalise denominator.)
+−=
5coti1
2
1 πk k = 1, 2, 3, 4. (AG)
Observes that original equation is a quartic with real coefficients, so roots occur in conjugate pairs and k = 0 must be rejected.
M1
A1
M1
A1
M1A1
A1
A1
B1
2
7 [9]
Page 7 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
7 Reduces M1 to echelon form. Finds. Dim(K1) Reduces M2 to echelon form. Finds Dim(K2) Obtains basis for K1. Obtains basis for K2, and shows
12KK ⊂ .
− 371834
9143
11412
4111
→…→
−
0000
0000
3210
4111
Dim(K1) = 4 – 2 =2 (AG)
−
0254
0143
1032
1111
→…→
−
−
0000
1000
3210
1111
(aef)
Dim(K2) = 4 – 3 = 1 (AG)
04 =+++ tzyx
032 =++− tzy Legitimately obtains:
Basis for K1 is
−
−
1
0
3
7
,
0
1
2
3
(OE)
032
0
=+−
=−++
tzy
tzyx
0=t Legitimately obtains:
Basis for K2 is
−
0
1
2
3
(OE) 12
KK ⊂⇒
M1A1
A1
A1
A1
M1
A1 A1
M1
A1
5
5
[10]
8 Forms AQE and solves. Writes CF. Correct form for PI and differentiates twice. Substitutes. Writes PI. Writes GS.
Uses y(0) = 5 to find A. Uses y′(0) = 1 to find B. Writes particular solution.
m2 + 2m + 5 = 0 ⇒m = –1 ± 2i CF: y = e–x (A cos 2x + B sin 2x) y = ke–2x ⇒ y′ = –2ke–2x ⇒ y″ = 4e–2x ⇒4k – 4k + 5k = 10 ⇒ k = 2 PI: y = 2e–2x GS: y = e–x (A cos 2x + B sin 2x) +2e–2x y = 5, x = 0⇒5 = A + 2⇒A = 3 y′ = –e–x (A cos 2x + B sin 2x) +e–x (–2A sin 2x + 2B cos 2x) –4e–2x y′ = 1, x = 0⇒1 = –3 + 2B – 4⇒B = 4 ∴ y = e–x (3 cos 2x + 4 sin 2x) +2e–2x
M1A1 A1
M1
M1 A1
A1
B1
M1 A1
A1
CAO 11 [11]
Page 8 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
9(i)
and
(ii)
Possible approach for
first two parts
together.
(i) can come after finding turning points: Continuous function (implied by graph)
⇒ (2,2.5) Max and (–1,1) Min
2
51 ≤≤⇒ y (AG)
N.B. Award B1 if Max and Min assumed without proof. i.e. 1/4 .
Writes 2
)1(1
2
3222
2
2
2
+
++=
+
++=
x
x
x
xxy
States 102
)1(2
2
≥⇒≥
+
+y
x
x
From this it is clear that (–1, 1) is a turning point.
Writes )2(2
)2(
2
5
2
3222
2
2
2
+
−−=
+
++=
x
x
x
xxy
States 2
50
)2(2
)2(2
2
≤⇒≥
+
−y
x
x
From this it is clear that (2, 22
1) is the other turning point.
(B1)
(B1)
(M1A1)
(B1)
(B1) (A1)
(M1)
(M1A1)
(A1)
(7)
(4)
Page 9 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
9 Forms quadratic equation in x.
Uses discriminant to obtain condition for real roots.
Differentiates and equates to zero.
Solves equation. States coordinates of turning points.
Expresses y in an appropriate form. (May alternatively divide numerator and denominator by x2. Finds y-intercept and intersection with y = 2. Completes graph.
322222
++=+ xxyyx 0)32(2)2( 2
=−+−−⇒ yxxy For real x 4 – 4(y – 2)(2y – 3) ≥ 0
0)1)(52( ≤−−⇒ yy
2
51 ≤≤⇒ y (AG)
0)322(2)24)(2(
0
22=++−++⇒
=′
xxxxx
y
⇒ (x –2)(x + 1) = 0⇒ x = –1 or x = 2
(Or substitutes y =1 and 2
5 in equation of C.)
Turning points are (–1, 1) and
2
12 ,2
2
122
2+
−+=
x
xy
As x → ±∞ y → 2 ∴ y = 2
Shows
2
11 ,0 and
2 ,
2
1
Completely correct graph.
M1 A1
M1
A1
M1
A1A1
M1
A1
B1
B1
4
3
2
2 [11]
Page 10 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
10
Differentiates and squares.
Uses formula for arc length.
Integrates and obtains value.
Uses formula for x-coordinate of centroid.
Integrates both expressions and obtains value.
Uses formula for y-coordinate of centroid.
Integrates both expressions and obtains value.
3)(
3
1 22
1
xyxy =′⇒=′
x
x
s d3
13
0 ∫ +=
)122(22243
12
3
0
2
3
−=−=
+=x
(AG)
∫
∫=
3
0
2
3
3
0
2
5
d33
2
d33
2
xx
xx
x
7
15
5
2
7
2
3
0
2
5
3
0
2
7
=
=
x
x
(= 2.14)
∫
∫ ×
=
3
0
2
3
3
0
3
d33
2
d27
4
2
1
xx
xx
y
8
5
5
2
33
2
427
2
3
0
2
5
3
0
4
=
=
x
x
(= 0.625)
B1
M1
A1A1
M1
A1A1 A1
M1
A1 A1
4
7
[11]
Page 11 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
11 EITHER
Note: (1) the parts can be either way round. (2) Insertion of limits in [ex sin x] causes the term to vanish.
∫ ∫+−== xxxxxIxxx
dcosecosedsine
∫−+−= xxxxxxx
dsinesinecose
∴2I = ex(sin x – cos x) π
π
0
0
)cos(sine2
1dsine∫
−=∴ xxxx
xx
2
e1
2
1
2
eππ +
=
−−= (AG)
∫=π
0
dsine xxInx
n
[ ] ∫ −
−=
ππ
0
1
0d)cossin(ee.sin xxxnx
nxxn
[ ]
−−+
−=
∫ −−
−
π
π
0
122
0
1
d)sinsinsin)1((cose
e.cossin0
xxxxnxn
xxn
nnx
xn
∫ −−+=−
π
0
22 dsincose)1(0n
nx
nIxxxnn (AG)
∫ −−−=−
π
0
22 dsin)sin1(e)1(n
nx
nIxxxnn
nnnInnInnIn )1()1()1(
2−−−=+∴
− ( )
2)1(1)1(
−
−=++−∴nnInnInnn
2
2 )1()1(−
−=+∴nnInnIn (AG)
135
10
6
26
20
26
20III ×==
( )π
π
e113
3
2
e1
13
6
5+=
+×=⇒ I
Mean value = )e1(13
3
0
dsine
0
5
π
π
ππ
+=
−
∫ xxx
M1
A1 M1
A1
M1
A1
A1
M1A1
A1
M1
A1
M1A1
4
6
4 [14]
Page 12 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
11
OR
Obtains direction of common perpendicular. Uses result for shortest distance between lines.
Solves equation.
Finds relevant vectors. Use of cross-product. Obtains shortest distance.
Finds 2nd vector in BCD (CD may already have been found.)
Finds normal vector to BCD. (Normal to ACD already found.)
Finds angle between planes = angle between normal vectors.
n =
110
114
−
−
m
kji
= –mi + 4(1– m)j + 4k
316)21(16
4
44
4
0
1
22
=++−+
−
−
−
mmm
m
m
⇒… 0440192
=+−⇒ mm
0)2)(219( =−−⇒ mm
⇒m = 2, since m is an integer. (AG)
CA =
− 4
0
1
and CD =
1
1
0
or AD =
−
5
1
1
−
−
=
− 1
1
4
17
1
401
11017
1
kji
17
18114
17
1 222=++ (= 1.03)
BC = 3i – j +5k
n =
110
513 −
kji
= –6i – 3j +3k ~ 2i + j – k
3
1
618
6
1141116
)kji2()kji4(cos ==
++++
−+⋅+−=θ
∴Angle between planes = cos–1
3
1 (AG)
M1A1
M1A1
A1
M1
A1
B1
M1
A1
B1
M1
M1
A1
7
3
4 [14]
Page 13 Mark Scheme: Teachers’ version Syllabus Paper
GCE A LEVEL – May/June 2012 9231 11
© University of Cambridge International Examinations 2012
11
OR
Alternatives for middle
part:
Or (a) Vector from D to any point on AC
Uses orthogonality to obtain t. Finds magnitude of perpendicular. Or (b) Finds length of AD (or CD) Finds projection of AD (or CD) onto AC. Finds perpendicular by Pythagoras.
−−
−
+
t
t
45
1
1
17
210
4
0
1
.
45
1
1
−=⇒=
−
−−
−
+
t
t
t
17
18114
17
1 222=++ (= 1.03)
27=AD
17
21
14
4
0
1
.
5
1
1
22
=+
−
−
17
18
17
44127 =− (= 1.03)
(B1)
(M1)
(A1)
(B1)
(M1)
(A1)
(3)
(3)